$y(x)$ is a solution for $y'(x) = f(x,y(x))$ defined for $x\in(a,b) , b<\infty$. $f(x,y)$ fulfill local Lipschitz condition with respect to $y$ for $x\in(a,b)$ (see below the condition as I was taught). So, $\lim_{x\rightarrow b} y(x) = L$ when $L$ is finale or $+-\infty$.
Lipschitz condition says: $\forall x_{0} \in (a,b) \exists\delta,c>0 : \forall y_1,y_2\in (y(x_0)-\delta,y(x_0)+\delta) , \forall x\in(x_0-\delta,x_0+\delta):$ $$ |f(x,y_2)-f(x,y_1)| \le |y_2 -y_1|c $$
My efforts so far: Assuming the limit doesn't exists, We get that $y(b)$ is defined (but is not a solution for the ODE) and $\exists \varepsilon_0 >0$ such that $\forall \delta >0 \exists x_\delta\in(b-\delta,b)$ such that:
$|y(x_\delta)-y(b)| >\varepsilon_0$ and $\forall x\in (b-\delta,b) |f(x)|<M$
Defining $\delta_n =1/n$ we get a sequence $\{x_n\}_1^\infty$ such that $\{x_n\} \rightarrow b$.
Using the derivative definition: $$\forall \lambda>0, \forall x_0\in(b-\lambda,b) \lim_{x\rightarrow x_0} \frac{y(x)-y(x_0)}{x-x_0} =f(x_0,y(x_0)) $$ so for $n,m\in\mathbb{N}$ we get: $$ \lim_{x\rightarrow x_n} \frac{y(x_n)-y(x_n)}{x-x_n} =f(x_n,y(x_n)) , \lim_{x\rightarrow x_m} \frac{y(x_m)-y(x_m)}{x-x_m} =f(x_m,y(x_m)) $$
Trying combining the last result with the assumption for Lipschitz condition: $$|f(x_n,y(x_n))-f(x_m,y(x_m))| \le |y(x_n) -y(x_m)|c$$ and using the last result, I am pretty much stuck, I can't find any idea how to extend the problem to $x=b$ , cause $y(x)$ is not a solution for $y'(x) = f(x,y(x))$ in $x=b$